Difference between revisions of "2004 AMC 10A Problems/Problem 7"

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==Problem==
 
==Problem==
A grocer stacks oranges in a pyramid-like stack whose rectangular base is 5 oranges by 8 oranges.  Each orange above the first level rests in a pocket formed by four oranges below.  The stack is completed by a single row of oranges.  How many oranges are in the stack?
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A grocer stacks oranges in a pyramid-like stack whose rectangular base is <math>5</math> oranges by <math>8</math> oranges.  Each orange above the first level rests in a pocket formed by four oranges below.  The stack is completed by a single row of oranges.  How many oranges are in the stack?
  
<math> \mathrm{(A) \ } 96 \qquad \mathrm{(B) \ } 98 \qquad \mathrm{(C) \ } 100 \qquad \mathrm{(D) \ } 101 {2}\qquad \mathrm{(E) \ } 134  </math>
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<math> \mathrm{(A) \ } 96 \qquad \mathrm{(B) \ } 98 \qquad \mathrm{(C) \ } 100 \qquad \mathrm{(D) \ } 101 \qquad \mathrm{(E) \ } 134  </math>
  
 
==Solution==
 
==Solution==
There are <math>5\times8=40</math> oranges on the 1st layer of the stack. When the 2nd layer is added on top of the first, it will be a layer of <math>4\times7=28</math> oranges.  When the third layer is added on top of the 2nd, it will be a layer of <math>3\times6=18</math> oranges, etc.
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There are <math>5\times8=40</math> oranges on the <math>1^{\text{st}}</math> layer of the stack. The <math>2^{\text{nd}}</math> layer that is added on top of the first will be a layer of <math>4\times7=28</math> oranges.  When the third layer is added on top of the <math>2^{\text{nd}}</math>, it will be a layer of <math>3\times6=18</math> oranges, etc.
  
Therefore, there are <math>5\times8+4\times7+3\times6+2\times5+1\times4=40+28+18+10+4=100</math> oranges in the stack <math>\Rightarrow\mathrm{(C)}</math>.
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Therefore, there are <math>5\times8+4\times7+3\times6+2\times5+1\times4=40+28+18+10+4=100</math> oranges in the stack. <math>\boxed{\mathrm{(C)}\ 100}</math>
  
==See Also==
 
  
*[[2004 AMC 10A Problems]]
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==Video Solution==
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https://youtu.be/H2OQPsSWG4A
  
*[[2004 AMC 10A Problems/Problem 6|Previous Problem]]
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Education, the Study of Everything
  
*[[2004 AMC 10A Problems/Problem 8|Next Problem]]
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== See also ==
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{{AMC10 box|year=2004|ab=A|num-b=6|num-a=8}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 00:14, 12 April 2022

Problem

A grocer stacks oranges in a pyramid-like stack whose rectangular base is $5$ oranges by $8$ oranges. Each orange above the first level rests in a pocket formed by four oranges below. The stack is completed by a single row of oranges. How many oranges are in the stack?

$\mathrm{(A) \ } 96 \qquad \mathrm{(B) \ } 98 \qquad \mathrm{(C) \ } 100 \qquad \mathrm{(D) \ } 101 \qquad \mathrm{(E) \ } 134$

Solution

There are $5\times8=40$ oranges on the $1^{\text{st}}$ layer of the stack. The $2^{\text{nd}}$ layer that is added on top of the first will be a layer of $4\times7=28$ oranges. When the third layer is added on top of the $2^{\text{nd}}$, it will be a layer of $3\times6=18$ oranges, etc.

Therefore, there are $5\times8+4\times7+3\times6+2\times5+1\times4=40+28+18+10+4=100$ oranges in the stack. $\boxed{\mathrm{(C)}\ 100}$


Video Solution

https://youtu.be/H2OQPsSWG4A

Education, the Study of Everything

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AMC 10 Problems and Solutions

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